: : Evaluate the entropy of mixing by mixing 1mole of A atoms
: : at P=1atm with 1mole of B atom at P=2atm .Assume that the
: : mixing is carried out at constant total volume.
: : (a) △S=   0
: : (b) △S= - Rln2/9
: : (c) △S=   3Rln2
: : (d) △S= - Rln2/3
: : (e) △S= - Rln1/3
: 假設 混合前兩氣體溫度相同(不假設這樣應該算不出題目要的答案)
: 混合前A體積2V , B體積V , 依題意混合後體積3V
: 因此混合後
: 氣體A 體積為原本的3/2倍,分壓變為原本的2/3倍
: ΔSa = ncv ln pf/pi + ncp ln Vf/Vi = ncv ln 2/3 + ncp ln 3/2
:                                    = n(ln 3/2)(cp-cv)
:                                    = R ln 3/2
: 氣體B 體積變為原本的3倍 ,分壓變為原本的1/3
: ΔSb = ncv ln pf/pi + ncp ln Vf/Vi = R ln 3
: => ΔS = ΔSa + ΔSb = R ln 9/2 = -R ln (2/9)
: Ans : (b)

  請問一下
  A和B原子混合不是應該要把混合Entropy考慮進去嗎?@@
  所以應該要加上ΔSm = -R(1+1)[(1/2)ln(1/2) + (1/2)ln(1/2)] = -Rln(1/4)
  因此答案是ΔS = ΔSa + ΔSb +ΔSm = -Rln(1/18)?


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